Simplify and expand the following expression: $ \dfrac{4}{z - 6}- \dfrac{5}{z - 4}- \dfrac{1}{z^2 - 10z + 24} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor the quadratic in the third term: $ \dfrac{1}{z^2 - 10z + 24} = \dfrac{1}{(z - 6)(z - 4)}$ Now we have: $ \dfrac{4}{z - 6}- \dfrac{5}{z - 4}- \dfrac{1}{(z - 6)(z - 4)} $ The least common multiple of the denominators is: $ (z - 6)(z - 4)$ In order to get the first term over $(z - 6)(z - 4)$ , multiply by $\dfrac{z - 4}{z - 4}$ $ \dfrac{4}{z - 6} \times \dfrac{z - 4}{z - 4} = \dfrac{4(z - 4)}{(z - 6)(z - 4)} $ In order to get the second term over $(z - 6)(z - 4)$ , multiply by $\dfrac{z - 6}{z - 6}$ $ \dfrac{5}{z - 4} \times \dfrac{z - 6}{z - 6} = \dfrac{5(z - 6)}{(z - 6)(z - 4)} $ Now we have: $ \dfrac{4(z - 4)}{(z - 6)(z - 4)} - \dfrac{5(z - 6)}{(z - 6)(z - 4)} - \dfrac{1}{(z - 6)(z - 4)} $ $ = \dfrac{ 4(z - 4) - 5(z - 6) - 1} {(z - 6)(z - 4)} $ Expand: $ = \dfrac{4z - 16 - 5z + 30 - 1}{z^2 - 10z + 24} $ $ = \dfrac{-z + 13}{z^2 - 10z + 24}$